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- 1. MechanicsTopic 2.2 Forces and Dynamics
- 2. Forces and Free-bodyDiagrams To a physicist a force is recognised by the effect or effects that it produces A force is something that can cause an object to Deform (i.e. change its shape) Speed up Slow Down Change direction
- 3. The last three of these can besummarised by stating that a forceproduces a change in velocityOr an acceleration
- 4. Free-body Diagrams A free-body diagram is a diagram in which the forces acting on the body are represented by lines with arrows. The length of the lines represent the relative magnitude of the forces. The lines point in the direction of the force. The forces act from the centre of mass of the body The arrows should come from the centre of mass of the body
- 5. Example 1 Normal/Contact ForceA block resting on a worktop Weight/Force due to Gravity
- 6. Example 2A car moving with a constant velocity Normal/Contact ForceResistance Motor Force Weight/Force due to Gravity
- 7. Example 3A plane accelerating horizontally Upthrust/Lift Motor Force Air Resistance Weight/Force due to Gravity
- 8. Resolving Forces Q. A force of 50N is applied to a block on a worktop at an angle of 30o to the horizontal. What are the vertical and horizontal components of this force?
- 9. Answer First we need to draw a free-body diagram 50N 30o
- 10. We can then resolve the force into the2 components Vertical = 50 sin 30o 50N 30o Horizontal = 50 cos 30o
- 11. Therefore Vertical = 50 sin 30o = 25N Horizontal = 50 cos 30o = 43.3 = 43N
- 12. Determining the ResultantForce Two forces act on a body P as shown in the diagram Find the resultant force on the body. 50N 30o 30N
- 13. Solution Resolve the forces into the vertical and horizontal componenets (where applicable) 50 sin 30o 50N 30o 30N 50 cos 30o
- 14. Add horizontal components and addvertical components. 50 sin 30o = 25N 50 cos 30o – 30N = 13.3N
- 15. Now combine these 2 components 25N R 13.3N R2 = 252 + 13.32 R = 28.3 = 28N
- 16. Finally to Find the Angle 25N R θ 13.3N tan θ = 25/13.3 θ = 61.987 θ = 62o The answer is therefore 28N at 62o upwards from the horizontal to the right
- 17. Springs The extension of a spring which obeys Hooke ´s law is directly proportional to the extending tension A mass m attached to the end of a spring exerts a downward tension mg on it and if it is stretched by an amount x, then if k is the tension required to produce unit extension (called the spring constant and measured in Nm-1) the stretching tension is also kx and so mg = kx
- 18. Spring Diagram x
- 19. Newton´s Laws The First Law Every object continues in a state of rest or uniform motion in a straight line unless acted upon by an external force
- 20. Examples Any stationary object! Difficult to find examples of moving objects here on the earth due to friction Possible example could be a puck on ice where it is a near frictionless surface
- 21. Equilibrium If a body is acted upon by a number of coplanar forces and is in equilibrium ( i.e. there is rest (static equilibrium) or unaccelerated motion (dynamic equilibrium)) then the following condition must apply The components of the forces in both of any two directions (usually taken at right angles) must balance.
- 22. Newton´s Laws The Second Law There are 2 versions of this law
- 23. Newton´s Second Law 1st version The rate of change of momentum of a body is proportional to the resultant force and occurs in the direction of the force. F = mv – mu F =∆ρ t t
- 24. Newton´s Second Law 2nd version The acceleration of a body is proportional to the resultant force and occurs in the direction of the force. F = ma
- 25. Linear Momentum The momentum p of a body of constant mass m moving with velocity v is, by definition mv Momentum of a body is defined as the mass of the body multiplied by its velocity Momentum = mass x velocity p = mv It is a vector quantity Its units are kg m s-1 or Ns It is the property of a moving body.
- 26. Impulse From Newtons second law F = mv – mu F =∆ρ t t Ft = mv – mu This quantity Ft is called the impulse of the force on the body and it is equal to the change in momentum of a body. It is a vector quantity Its units are kg m s-1or Ns
- 27. Law of Conservation of LinearMomentum The law can be stated thus When bodies in a system interact the total momentum remains constant provided no external force acts on the system.
- 28. Deriving This Law To derive this law we apply Newton´s 2nd law to each body and Newton´s 3rd law to the system i.e. Imagine 2 bodies A and B interacting If A has a mass of mA and B has a mass mB If A has a velocity change of uA to vA and B has a velocity change of uB to vB during the time of the interaction t
- 29. Then the force on A given by Newton 2 isFA = mAvA – mAuA tAnd the force on B isFB = mBvB – mBuB tBut Newton 3 says that these 2 forces areequal and opposite in direction
- 30. Therefore mAvA – mAuA = -(mBvB – mBuB) t tTherefore mAvA – mAuA = mBuB – mBvBRearranging mAvA + mBvB = mAuA + mBuBTotal Momentum after =Total Momentum before
- 31. Newton´s Laws The Third Law When two bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A.
- 32. Example of Newton´s 3rd Q. According to Newton’s third Law what is the opposite force to your weight? A. As your weight is the pull of the Earth on you, then the opposite is the pull of you on the Earth!
- 33. Newton´s 3rd Law The law is stating that forces never occur singularly but always in pairs as a result of the interaction between two bodies. For example, when you step forward from rest, your foot pushes backwards on the Earth and the Earth exerts an equal and opposite force forward on you. Two bodies and two forces are involved.
- 34. Important The equal and opposite forces do not act on the same body!

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